Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $r = \dfrac{k^3 + 3k^2 - 10k}{k^3 + 8k^2 + 15k} \div \dfrac{-3k + 30}{3k + 9} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{k^3 + 3k^2 - 10k}{k^3 + 8k^2 + 15k} \times \dfrac{3k + 9}{-3k + 30} $ First factor out any common factors. $r = \dfrac{k(k^2 + 3k - 10)}{k(k^2 + 8k + 15)} \times \dfrac{3(k + 3)}{-3(k - 10)} $ Then factor the quadratic expressions. $r = \dfrac {k(k + 5)(k - 2)} {k(k + 5)(k + 3)} \times \dfrac {3(k + 3)} {-3(k - 10)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac { k(k + 5)(k - 2) \times 3(k + 3)} { k(k + 5)(k + 3) \times -3(k - 10)} $ $r = \dfrac {3k(k + 5)(k - 2)(k + 3)} {-3k(k + 5)(k + 3)(k - 10)} $ Notice that $(k + 5)$ and $(k + 3)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {3k\cancel{(k + 5)}(k - 2)(k + 3)} {-3k\cancel{(k + 5)}(k + 3)(k - 10)} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $r = \dfrac {3k\cancel{(k + 5)}(k - 2)\cancel{(k + 3)}} {-3k\cancel{(k + 5)}\cancel{(k + 3)}(k - 10)} $ We are dividing by $k + 3$ , so $k + 3 \neq 0$ Therefore, $k \neq -3$ $r = \dfrac {3k(k - 2)} {-3k(k - 10)} $ $ r = \dfrac{-(k - 2)}{k - 10}; k \neq -5; k \neq -3 $